Problem: $f(x) = \dfrac{ 3 }{ \sqrt{ 6 - \lvert x \rvert } }$ What is the domain of the real-valued function $f(x)$ ?
Solution: First, we need to consider that $f(x)$ is undefined anywhere where the radicand (the expression under the radical) is less than zero. So we know that $6 - \lvert x \rvert \geq 0$ This means $\lvert x \rvert \leq 6$ , which means $-6 \leq x \leq 6$ Next, we need to consider that $f(x)$ is also undefined anywhere where the denominator is zero. So we know that $\sqrt{ 6 - \lvert x \rvert } \neq 0$ , so $\lvert x \rvert \neq 6$ This means that $x \neq 6$ and $x \neq -6$ So we have four restrictions: $x \geq -6$ $x \leq 6$ $x \neq -6$ , and $x \neq 6$ Combining these four, we know that $x > -6$ and $x < 6$ ; alternatively, that $-6 < x < 6$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid -6< x <6\, \}$.